In a local frame of $E$, a <connection on a vector bundle> has the form
$$
d^{\mathcal A}=d+A\wedge,
$$
where $A$ is a matrix of one-forms. Under a frame change $g$, its matrix transforms as
$$
A'=g^{-1}Ag+g^{-1}dg.
$$
The <covariant exterior derivative> on an $E$-valued $r$-form $\sigma$ is
$$
d^{\mathcal A}\sigma=d\sigma+A\wedge\sigma.
$$
The <curvature form of a connection> is
$$
F=dA+A\wedge A.
$$
Using the graded Leibniz rule,
$$
(d^{\mathcal A})^2\sigma
=d(A\wedge\sigma)+A\wedge d\sigma+A\wedge A\wedge\sigma
=(dA+A\wedge A)\wedge\sigma
=F\wedge\sigma.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 115 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-115/3/a/solution